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damaskus [11]
3 years ago
13

Yo if u need an bralest i got chu

Mathematics
2 answers:
Alik [6]3 years ago
8 0

Answer:

yes

Step-by-step explanation:

dybincka [34]3 years ago
4 0

Answer:

PLease

Step-by-step explanation:

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Evaluate the summation from n equals 1 to infinity of the quotient of 3 and 2 raised the quantity n minus 1 power
Shalnov [3]

Answer:

3*∑ (1/2)^n = 6

Step-by-step explanation:

We have the summation from n = 1 to n = ∞, for:

∑ 3/(2)^(n - 1)

We know that the summation between k = 0, and k = N - 1 for:

∑ r^k = (1 - r^N)/(1 - r)

if we have the summation between k = 0 and k = ∞ - 1 = ∞

(here we used that ∞ is really big, then ∞ - 1 = ∞)

In the numerator we will have the term r^∞, if 0 < r < 1, then r^∞ = 0.

Then if we assume that 0 < r < 1 we can write:

∑ r^k =  1/(1 - r)

In our case, we can rewrite our summation as:

3*∑ (1/2)^n

for n = 0 to  n = ∞

You can see that i changed the limits for n, this does not really matter because we are summing between a number and infinity, the only thing you need to take care is that now the power is n, instead of (n - 1)

Then we have r = (1/2) which is clearly smaller than 1.

then (1/2)

Then this summation is equal to:

3*∑ (1/2)^n = 3*( 1/(1 - 1/2)) = 3*( 1/( 2/2 - 1/2)) = 3*( 1/(1/2)) = 3*2 = 6

3*∑ (1/2)^n = 6

5 0
3 years ago
Okay I got a question
Maslowich
What’s the question?
7 0
3 years ago
Jake went to the zoo
kkurt [141]

Answer:

i got u he was eaten by a bear.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Drag the blue point to a place on the number line
sergejj [24]

Answer:

The number line is shown below.

Step-by-step explanation:

We need to represent number that is less than 3 on a number line.

Let x be numbers.

So, x.

Now, we have a number which represents x<3. Here 3 is not included in the solution set, so there is an open circle on 3 and left side of 3 are in the solution set.

The number line is shown below.

4 0
3 years ago
Which expression is equivalent to (sqrt3 + i)^3?
Anarel [89]

Express √3 + i in polar form:

|√3 + i| = √((√3)² + 1²) = √4 = 2

arg(√3 + i) = arctan(1/√3) = π/6

Then

√3 + i = 2 (cos(π/6) + i sin(π/6))

By DeMoivre's theorem,

(√3 + i)³ = (2 (cos(π/6) + i sin(π/6)))³

… = 2³ (cos(3 • π/6) + i sin(3 • π/6))

… = 8 (cos(π/2) + i sin(π/2))

… = 8i

4 0
2 years ago
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